Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm trying to solve $$\sin(x)\frac{d}{dx}\beta \left ( x \right )+\cos(x)\beta (x)=1$$ What I get is : $$\beta (x)=\beta \left ( \alpha \right )e^{\sin(\alpha )-\sin(x)}+e^{-\sin(x)}\int_{\alpha }^{x}e^{\sin(t)}dt$$ But I think that this solution is incorrect .The textbook says that there's exactly one solution that has a finite limit as $x$ tends to $0$ . But all the solutions I get have a finite limit . So what's the correct solution?

share|improve this question
    
Note that you can use \sin, \cos for the $\sin,\cos$ functions. see the difference between $sin(x)$ and $\sin(x)$ –  Jean-Sébastien Nov 3 '12 at 20:22
    
Unless $\beta(x)$ is a special function, I see that since $sin^2(x)+cos^2(x)=1$ which makes $\beta(x)=cos(x)$ is a solution. –  Emmad Kareem Nov 3 '12 at 20:29
1  
@EmmadKareem That was my first thought, but it gives $-\sin^2(x)+\cos^2(x)$, notice the "negative sign" (pun!) –  Logan Nov 3 '12 at 20:44
    
@Logan, thanks, I need thicker glasses and a new brain :) –  Emmad Kareem Nov 3 '12 at 20:54
add comment

1 Answer

$\sin(x) \beta'(x) + \cos(x) \beta(x) = \sin(x) \beta'(x) + (\sin(x))' \beta(x) = 1$. Hence, $$(\sin(x) \beta(x))' = 1 \implies \sin(x) \beta(x) = x + c \implies \beta(x) = \dfrac{x+c}{\sin(x)}$$ Now as $x \to 0$, we want $\lim_{x \to 0} \beta(x)$ to exist i.e. $\lim_{x \to 0} \dfrac{x+c}{\sin(x)}$ exists i.e. $$\lim_{x \to 0} \dfrac{x+c}{\sin(x)} = 1 + \lim_{x \to 0} \dfrac{c}{\sin(x)} \,\,\,\,\,\,\, \text{exists}.$$ Hence, $c=0$ and therefore $\beta(x) = \dfrac{x}{\sin(x)}$.

EDIT

To proceed through your way, we have that $$\beta'(x) + \cot(x) \beta(x) = \csc(x)$$ Hence, the integrating factor $I(x) = \exp \left(\displaystyle \int M(y) dy\right) = \exp(\log(\sin(x))) = \sin(x)$. Hence, the solution is $$\beta(x) \sin(x) = \int \csc(y) \sin(y) dy = x + c$$

share|improve this answer
    
You can also finish it by observing that if $\lim_{x \to 0} \dfrac{x+c}{\sin(x)}$ exists then, $\lim_{x \to 0} \dfrac{x+c}{\sin(x)}\cdot \sin (x)=0$. –  N. S. Nov 3 '12 at 20:26
    
Thanks Marvis , But what about my solution ,Is it wrong? –  Nabil Nov 3 '12 at 20:27
    
@Nabil Did you check if it satisfies the equation? I think it doesn't... –  N. S. Nov 3 '12 at 20:29
    
Yes ,I tried but I get a complicated expression . I think that my solution is wrong . –  Nabil Nov 3 '12 at 20:35
    
But $cos(x)$ is a solution that's finite as $x$ tends to $0$ . So your solution is not the most general right? –  Nabil Nov 3 '12 at 20:41
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.